 | 1801 - 658 Seiten
...the street. • t Ans. 76- 1 2333 35 feet. PROBLEM IV. 7o f:nd tlie area of a trapezoid. • RULE.* Multiply the sum of the two parallel sides by the...between them, and half the product will be the area. • EXAMPLES. * DEMONSTRATION. or (because B»=DE) =-, .-. A ABD+ A BCD, or , , , ABxDE , DCxDE the... | |
 | Abel Flint - 1804 - 226 Seiten
...8925X0.47076=4201 the double Area of the Triangle. PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the Sum of the two parallel Sides by half the perpendicular distance ; the Product will be the Area.... | |
 | James Thompson - 1808 - 176 Seiten
...trafiezoid, or quadrangle, <u'o cf •whose opposite sides are parallel. RULE — Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product •will be the area. EXAMPLES. 13. Required the area of a trapezoid whose parallel sidef »re 25 feel 6 inches, and 18 feet... | |
 | Samuel Webber - 1808 - 466 Seiten
...breadth of the street. Ans. 76' 1233335 feet. 414PR0BLEM IV. To find the area of a trapezold. RULE.* Multiply the sum of the two parallel sides by the perpendicular distance between them, and hall" the product will be the area. EXAMPLES. 1. In a trapezoid, the parallel sides are AB 7' 5, and... | |
 | Abel Flint - 1808 - 190 Seiten
...the double Area of the Triangle. • PROBLEM X. To find the Area of a Trapezoid. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the sum of the two parallel Sides by half the perpendicular distance ; the Product will be the Area.... | |
 | Thomas Keith - 1817 - 306 Seiten
...acres. <• 2 roods 21 perches. PROBLEM VIII. • To find the Area of a Trapezoid. RULE *. Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will be the area. Example 1. Let AB c D JE. be a trapezoid, the side '-. A )•. — 23,... | |
 | Anthony Nesbit, W. Little - 1822 - 916 Seiten
...the grain? Ant. 97.3383 bushels. PROBLEM VII. To Jind the area of a trapezoid. RULE. • By the Pen. Multiply the sum of the two parallel sides by the...between them, and half the product will be the area in square inches. Divide this area by 2 82, 231, and 2150.42, and the respective quotients will be... | |
 | Anthony Nesbit - 1824 - 476 Seiten
...Ans. 1131^.2 in. 9 pa. PROBLEM VIII. To find the area of a Irapezoid. RULE. Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. Or, half the sum of the sides multiplied by their distance will give the area. EXAMPLE. 1. What is... | |
 | John Nicholson - 1825 - 838 Seiten
...when the squre of AB has been subtracted. 63 I 189 3 1 189 Prot. 4. To find the Area ofaTrapezoid. Multiply the sum of the two parallel sides by the...between them, and half the product will be the area. Ex. In a trapezoid, the parallel sides are AB 7, and CD 12, and the perpendicular distance AP or CN... | |
 | Abel Flint - 1825 - 252 Seiten
...0.47076=4201 tbe double Area of the Triangle. PROBLEM X. To find the Jbeaof a TrapezoiA. RULE. Multiply half the Sum of the two parallel Sides by the perpendicular distance between them, or the sum of the two parallel Sides by half the perpendicular distance, the product will be the Area.... | |
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Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. 
